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Edexcel M4 Exam Discussion Thread [June 5th]

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The one where you are required to find the speed of B.

If it is then I got that question completely wrong. Think I dropped all the marks but 1. Still ended up with an A* on the paper though because the boundary was so low :smile:

Reply 121

simonli2575

Original post by Gome44

If it is then I got that question completely wrong. Think I dropped all the marks but 1. Still ended up with an A* on the paper though because the boundary was so low :smile:

Ah no, that one was the easy one. I was talking about the one in which boat B is moving due east and some time later, B is 2sqrt(3) km due south of A.

Reply 122

Gome44

Original post by simonli2575

Ah no, that one was the easy one. I was talking about the one in which boat B is moving due east and some time later, B is 2sqrt(3) km due south of A.

Oh, I found the difficulty the other way around. I think on the one I sent a photo of I thought P was moving so got that completely wrong. For the one you're talking about though, the angles were like 150 and 210 so it was screaming for I and j vectors

Reply 123

ThatPerson

Original post by Vesniep

Me too(M3 90,Fp2 100). How did you do in FP2, M3? M4 is not the best unit to rely on (you can aim for A* but not be sure)
If you have M2 and/or FP3 it is better to think of getting a good mark in these units,in my opinion( they are safer)

I'm fairly sure I got 95+ in FP2, and somewhere between 80-90 in M3 depending on the boundaries, so if I take the minimums, to guarantee an A* tomorrow I would need 95 UMS.

Reply 124

Gome44

Hoping for a paper like last years (mostly easy but some tricky parts so the grade boundaries fall)

Reply 125

ThatPerson

What is the reason for why the component of the velocity that is perpendicular to the common remains unchanged? Also, why does the fact that all collisions in M4 are smooth mean that the mutual reaction acts along the common normal at the point of impact?

If we were solving a problem involving collisions of rough elastic bodies, what would change?

Reply 126

Gawain

Original post by ThatPerson

When two objects collide, they exert a force upon each other perpendicular to their plane of contact (along the normal of the plane).
This means an impulse is only felt parallel to the line of centres and therefore the velocities perpendicular to this line are unchanged (impulse is 0).

If the objects are completely smooth then there is no force perpendicular to the normal. If they were rough there would be a component frictional force perpendicular to the normal and so both components of velocity would change rather than just one.

Reply 127

ThatPerson

Original post by Gawain

Thanks for the explanation. So if I have the following question:

"A small smooth spherical ball of mass m falls vertically at strikes a fixed smooth inclined plane with speed u.

(a) Explain why the component of the velocity of the ball parallel to the plane is not affected by the impact"

The answer would be something like "When the collision occurs the ball and plane exert a force on each other that is perpendicular to the plane, hence there is no impulse parallel to the plane and so the parallel velocity is unchanged"?

Reply 128

TeeEm

quick question on the relative motion (M4 edexcel)
only a few minutes old ..

I will post the solution real soon

M4 relative velocity.pdf14.3KB

Reply 129

simonli2575

Original post by TeeEm

quick question on the relative motion (M4 edexcel)
only a few minutes old ..

I will post the solution real soon

How about a question using relative velocities to find the minimum separation given A is trying go get away from B?

Reply 130

TeeEm

Original post by simonli2575

How about a question using relative velocities to find the minimum separation given A is trying go get away from B?

will try but I do not think it will fit with my to do list before the M4 exam.

Reply 131

TeeEm

Original post by TeeEm

quick question on the relative motion (M4 edexcel)
only a few minutes old ..

I will post the solution real soon

Question with solution from post 129

M4 relative velocity with solution.pdf177.6KB

Reply 132

jjpneed1

Can someone please explain question 3 June 2014

Why is 1/2 (x^2 + y^2) = x^2 + y^2/9 ??? If the speed is half then surely

1/2 \sqrt{x^2 + y^2} = \sqrt{x^2 + y^2/9}

so 1/4(x^2 + y^2) = x^2 + y^2/9

Am I being really dumb or wtf??

Reply 133

CowsWhoStare

Can someone explain to me how to find the least value of speed for something to intercept something else? like in June 2013 question 5

Posted from TSR Mobile

Reply 134

Gawain

Original post by ThatPerson

Yeah that looks good.

Original post by CowsWhoStare

Can someone explain to me how to find the least value of speed for something to intercept something else? like in June 2013 question 5

Posted from TSR Mobile

To find the minimum velocity for interception, the relative velocity (which is in the direction of the initial distance) is perpendicular to the 'interceptor's' velocity.

For questions relating to crossing a river, the quickest way is to swim straight across (perpendicular to the bank).

Original post by TeeEm

Question with solution from post 129

This is just a stylistic question but do you think it is better to separate your diagrams or combine them like this:
ImageUploadedByStudent Room1433441016.865979.jpg

ImageUploadedByStudent Room1433441016.865979.jpg

(edited 8 years ago)

Reply 135

Gawain

Original post by jjpneed1

Can someone please explain question 3 June 2014

Why is 1/2 (x^2 + y^2) = x^2 + y^2/9 ??? If the speed is half then surely

1/2 \sqrt{x^2 + y^2} = \sqrt{x^2 + y^2/9}

so 1/4(x^2 + y^2) = x^2 + y^2/9

Am I being really dumb or wtf??

No that's right,
If you follow on from yours the ratio you get is y/x=(27/5)^0.5 which is what the final line has.

The ratio isn't 3/(7^0.5).

Reply 136

TeeEm

Original post by Gawain

Yeah that looks good.

To find the minimum velocity for interception, the relative velocity (which is in the direction of the initial distance) is perpendicular to the 'interceptor's' velocity.

For questions relating to crossing a river, the quickest way is to swim straight across (perpendicular to the bank).

This is just a stylistic question but do you think it is better to separate your diagrams or combine then like this:
ImageUploadedByStudent Room1433441016.865979.jpg

everybody has their style and this particular type of question has way too many styles ...
Apologies to any student which finds my style difficult to follow, however feel free to attach an alternative solution if it is going to benefit some students

Reply 137

Gawain

Original post by TeeEm

I don't find your style confusing, to the contrary I think it's clearer. It's just mark schemes combine there diagrams so I thought that's what is expected.

Reply 138

TeeEm

Original post by Gawain

I don't find your style confusing, to the contrary I think it's clearer. It's just mark schemes combine there diagrams so I thought that's what is expected.

I am not ashamed to admit I cannot fully follow diagrams in marking schemes (or indeed some of the diagrams in the examples in the new book of EDEXCEL).
The way I present it is the way I understood the topic as a learner a very long time ago. (Fashions do change)

Reply 139

jjpneed1

Original post by Gawain

No that's right,
If you follow on from yours the ratio you get is y/x=(27/5)^0.5 which is what the final line has.

The ratio isn't 3/(7^0.5).

Using dot product on the velocity before and after gives the wrong answer for some reason though, but yeah the ratio is still right.

Also another problem with Q5 on the same paper

In the conservation of momentum calculation I used x and y going in the opposite direction to that shown in the diagram in the mark scheme, so

-u = x + 3y and x - y = u/3

but this gives x=0, y=-u/3 which is wrong... why would it matter which why I label the velocities...? Somehow I went through the other mechanics modules without encountering this problem

EDIT: nvm sign error using restitution I believe

(edited 8 years ago)